Magnetic Force on a Charged Particle Magnetic Force

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-Magnetic Force on a Charged Particle -Magnetic Force on a Current-Carrying Wire -Torque on

-Magnetic Force on a Charged Particle -Magnetic Force on a Current-Carrying Wire -Torque on a Current-Carrying Loop Fermi Lab, Chicago Illinois Circumference 6. 3 km Mass Spectrometer DC Motor AP Physics C Mrs. Coyle

 • A magnetic field can exert a force on a charged particle that

• A magnetic field can exert a force on a charged particle that moves in it.

Three ways we talk about “magnetic field”. • Magnetic Field: Regions surrounding a magnet

Three ways we talk about “magnetic field”. • Magnetic Field: Regions surrounding a magnet where another magnet or a moving electric charge will feel a force of attraction or repulsion. • Magnetic Field Lines: exit the north pole and enter the south. • Magnetic Field Strength, B • Vector, Unit: Tesla, T • Named after Nikola Tesla

Poles • Law of Poles: Like poles repel, unlike poles attract. • The force

Poles • Law of Poles: Like poles repel, unlike poles attract. • The force between two poles varies as the inverse square of the distance between them. • A single pole (monopole) has not been isolated.

Force on a Charged Particle Moving in a Magnetic Field -Magnetic fields only exert

Force on a Charged Particle Moving in a Magnetic Field -Magnetic fields only exert forces on moving charged particles or other magnets. • F = |q|(v x B) = |q|v. B sin q • vector cross-product • q is the angle between v and B • F=0 for q =0 or 1800

Remember: Cross Product Using Determinants or

Remember: Cross Product Using Determinants or

Remember: Properties of Cross Products of Unit Vectors ixi=0 jxj=0 kxk=0 ixj=k jxk=i kxi=j

Remember: Properties of Cross Products of Unit Vectors ixi=0 jxj=0 kxk=0 ixj=k jxk=i kxi=j

To find the direction of the force use the Right Hand Rule For a

To find the direction of the force use the Right Hand Rule For a positive test charge: Thumb v Fingers B Out of palm F The force is always perpendicular to the v. B plane

For a negative particle the F is opposite to what it would be for

For a negative particle the F is opposite to what it would be for a positive particle (use left hand)

Alternate Rule: Right Hand Curl Rule • Curl fingers from v to B F

Alternate Rule: Right Hand Curl Rule • Curl fingers from v to B F = q(v x B) • F is in the direction of the thumb • Similarly used for the direction of torque (t=r x F)

Graphical Representation of the Magnetic Field Vector (Strength), B x field lines pointing into

Graphical Representation of the Magnetic Field Vector (Strength), B x field lines pointing into the page ● field lines pointing out of the page

Question • Find the direction of the magnetic force acting on the +charged particle

Question • Find the direction of the magnetic force acting on the +charged particle entering the magnetic field with a velocity v perpendicular to B. x x V x x x Answer: Upwards x x x

What motion will the particle in the previous example undergo (particle entered the Bfield

What motion will the particle in the previous example undergo (particle entered the Bfield in a direction perpendicular to B? • Circular Motion • Magnetic force will represent the centripetal force • http: //online. cctt. org/physicslab/content/ap plets/Java. Phys. Math/java/partmagn/index. html

Examples: http: //physicslearning. colorado. edu/Pira. Home/Physics. Drawings. htm

Examples: http: //physicslearning. colorado. edu/Pira. Home/Physics. Drawings. htm

Does the magnetic force do work? • F is always perpendicular to the displacement

Does the magnetic force do work? • F is always perpendicular to the displacement • F can change the direction of v not the magnitude • F cannot do work, cannot change KE

Mass Spectrometer U. S. Department of Energy http: //doegenomestolife. org

Mass Spectrometer U. S. Department of Energy http: //doegenomestolife. org

Motion of Alpha- Beta-Gamma Particles in a Magnetic Field 1) Alpha particles, positive helium

Motion of Alpha- Beta-Gamma Particles in a Magnetic Field 1) Alpha particles, positive helium nuclei, charge +2 e 2) Gamma rays, (no charge) electromagnetic radiation 3) Beta particles, electrons charge -1 e

Problem 1 A proton is accelerated through a constant electric field (parallel plates) and

Problem 1 A proton is accelerated through a constant electric field (parallel plates) and acquires kinetic energy of 4 e. V. It enters perpendicularly to the 2 T field of a detector as shown. Charge of a proton=1. 6 x 10 -19 C, mass of proton=1. 67 x 10 -27 kg, ignore gravity. a) Draw the path of the positive charge as it enters the magnetic field.

Problem 1 cont’d b) Calculate the force acting on the charge due to the

Problem 1 cont’d b) Calculate the force acting on the charge due to the magnetic field. Charge of a proton=1. 6 x 10 -19 C , mass of proton=1. 67 x 10 -27 kg KE= 4 e. V B= 2 T Ans: v= 2. 77 x 104 m/s, F= 8. 86 x 10 -15 N

Problem 1 cont’d c) Calculate the distance on the detector where the particle will

Problem 1 cont’d c) Calculate the distance on the detector where the particle will land (radius of the circular path). Ignore gravity. Charge of a proton=1. 6 x 10 -19 C, mass of proton=1. 67 x 10 -27 kg KE= 4 e. V B= 2 T Ans: 1. 45 x 10 -4 m

Large Hadron Collider CERN (Conseil Européen pour la Recherche Nucléaire) Switzerland- France 2008 Circumference:

Large Hadron Collider CERN (Conseil Européen pour la Recherche Nucléaire) Switzerland- France 2008 Circumference: 27 km

 • Particle Accelerators use electric and magnetic fields to accelerate charged particles. •

• Particle Accelerators use electric and magnetic fields to accelerate charged particles. • Cyclotron Applet

Magnetic Force on a Current Carrying Wire F = I (L x B) =

Magnetic Force on a Current Carrying Wire F = I (L x B) = I L B sinq • I is the current • L is a vector of magnitude of the length of the wire and direction that of the current • q is the angle formed between I and B What is the direction of the magnetic force acting on this wire? F

The force acting on a wire of arbitrary shape is the same as if

The force acting on a wire of arbitrary shape is the same as if it were a straight wire with the same ends • The total force is:

What is the net force acting on this current-carrying loop?

What is the net force acting on this current-carrying loop?

Which way will a loop turn?

Which way will a loop turn?

Forces on a Current-Carrying Loop For B as shown: F 1 = F 3

Forces on a Current-Carrying Loop For B as shown: F 1 = F 3 = 0 F 2 = F 4 = Ia. B

Torque Acting on the Current-Carrying Loop F 2 = F 4 = Ia. B

Torque Acting on the Current-Carrying Loop F 2 = F 4 = Ia. B Torque acting on the loop: Top View

Torque Acting on the Current. Carrying Loop

Torque Acting on the Current. Carrying Loop

Magnetic Force on a Current Carrying Wire • Lorentz Force- Magnetic Force on a

Magnetic Force on a Current Carrying Wire • Lorentz Force- Magnetic Force on a Current Carrying Wire (Fendt Applet)

DC Motor

DC Motor

DC Motor

DC Motor

Problem #9 A proton moves with a velocity of v=(2 i-4 j+k)m/s in a

Problem #9 A proton moves with a velocity of v=(2 i-4 j+k)m/s in a region in which the magnetic field is B=(i+2 j-3 k)T. What is the magnitude of the magnetic force this charge experiences? Ans: 2. 34 x 10 -18 N

Problem #13 A wire 2. 80 m in length carries a current of 5.

Problem #13 A wire 2. 80 m in length carries a current of 5. 00 A in a region where a uniform magnetic field has a magnitude of 0. 390 T. Calculate the magnitude of the magnetic force on the wire assuming the angle between the magnetic field and the current is (a) 60. 0°, (b) 90. 0°, (c) 120°. Ans: a)4. 73 N, b)5. 46 N, c)4. 73 N

Problem # 21 A small bar magnet is suspended in a uniform 0. 250

Problem # 21 A small bar magnet is suspended in a uniform 0. 250 -T magnetic field. The maximum torque experienced by the bar magnet is 4. 60 × 10– 3 N · m. Calculate the magnetic moment of the bar magnet. Ans: 18. 4 m. A m 2

Problem #54 A 0. 200 -kg metal rod carrying a current of 10. 0

Problem #54 A 0. 200 -kg metal rod carrying a current of 10. 0 A glides on two horizontal rails 0. 500 m apart. What vertical magnetic field is required to keep the rod moving at a constant speed if the coefficient of kinetic friction between the rod and rails is 0. 100? Ans: F=0. 196 N, B=0. 039 T